**
A sporting goods store sells 90 ski jackets in a season for $275 each.
Each $15 decrease in the price results in five more jackets being
sold. What is the lowest price that would produce revenues of at least
$19 600? How many jackets would be sold at this price?
**
Let x = each $15 decrease in price.
new price = 275-15x
new # of ski jackets sold = 90+5x
Let y = Revenue of $19 600
y = (275-15x)(90+5x)
19600 = (275-15x)(90+5x)
19600 = 24750 + 1375x - 1350x - 75x^2
19600 =-75x^2 + 25x + 24750
0 = -75x^2 + 25x + 5150
This is where I'm stuck and don't know what to do. Can anyone help me out and show me what I'm suppose to do next?
Your working appears to be correct so far. Divide both sides by $-75$. This makes the equation look like this:
$$x^2-\frac{1}{3}x-\frac{206}{3}=0\implies x^2-\frac{1}{3}x=\frac{206}{3}$$
Now, we will do something which is usually called "completing the square". It is essentially the way you prove the quadratic formula.
\begin{align*} x^2-\frac{1}{3}x+\frac{1}{36} &= \frac{206}{3}+\frac{1}{36} \\ \implies \left(x-\frac{1}{6}\right)^2 &= \frac{2473}{36} \\ \implies x-\frac{1}{6} &= \pm\frac{\sqrt{2473}}{6} \\ \implies x &= \frac{1\pm\sqrt{2473}}{6} \end{align*}
Another way to get to this stage would be to use "the quadratic formula". If yo have an equation $ax^2+bx+c=0$ that needs solving, the quadratic formula states that the solutions are:
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Directly using your equation: $-75x^2+25x+5150=0$, so $a=-75,b=25,c=5150$. If yo plug into the formula, you find:
$$x=\frac{-25\pm\sqrt{25^2-4(-75)5150}}{2(-75)}$$
and this turns out to be the same as $x = \frac{1\pm\sqrt{2473}}{6}$.
Note that the solution $x=\frac{1-\sqrt{2473}}{6}$ is negative, which will result in putting the jackets at a higher prices: this isn't what the problem is asking for. You will need to use $x=\frac{1+\sqrt{2473}}{6}$.
I'm sure you can finish it off from here.